I'm getting a bit confused here. I keep seeing two common answers for these types of situations: $51\choose5$ and $52\choose4$. Which of these answer which question? Are there $51\choose5$ hands containing the ace of clubs, or are there $52\choose4$ (or neither)?
Likewise, are there $51\choose5$ hands not containing the ace of clubs, or are there $52\choose4$?
I'm really hoping one value answers one question, and the other value answers the other question, otherwise I'll be even more confused...
I assume you're taking about $5$-card (perhaps poker) hands.
To get a hand without the ace of clubs, you need to choose $5$ from the other $51$ cards, and there are $\binom{51}5$ ways to do that.
To get a hand with the ace of clubs, you need to choose the remaining $4$ cards from the other $51$ cards, and there are $\binom{51}4$ ways to do that.
The sum is $\binom{51}5+\binom{51}4=\binom{52}5$, as it must be.